CBSE Class 10 Maths HOTs Arithmetic Progressions Set D

Short Answer Type Questions

Question. Find the 12th term from the end of the AP -2, -4, -6, ..., -100.
Answer: From end: \( a = -100, d = +2 \).
\( a_{12} = -100 + (12-1)2 = -100 + 22 = -78 \).

Question. How many numbers lie between 10 and 300, which divided by 4 leave a remainder 3?
Answer: Sequence: 11, 15, 19, ..., 299.
\( a = 11, d = 4, a_n = 299 \).
\( 11 + (n-1)4 = 299 \Rightarrow 4(n-1) = 288 \Rightarrow n-1 = 72 \Rightarrow n = 73 \).

Question. If m times the mth term of an AP is equal to n times its nth term, show that the (m + n)th term of the AP is zero.
Answer: \( m[a+(m-1)d] = n[a+(n-1)d] \)
\( ma + m(m-1)d = na + n(n-1)d \)
\( a(m-n) + [m^2-m-n^2+n]d = 0 \)
\( a(m-n) + [(m-n)(m+n) - (m-n)]d = 0 \)
Dividing by \( m-n \): \( a + (m+n-1)d = 0 \), which is the \( (m+n) \)th term.

Question. Find the sum of first 20 terms of the following AP sequence 1, 4, 7, 10, ...
Answer: \( a=1, d=3, n=20 \).
\( S_{20} = \frac{20}{2}[2(1) + (19)3] = 10[2 + 57] = 10 \times 59 = 590 \).

Question. Which term of the AP : 120, 116, 112, ... is first negative term?
Answer: \( a=120, d=-4 \). Let \( a_n < 0 \).
\( 120 + (n-1)(-4) < 0 \Rightarrow 120 - 4n + 4 < 0 \Rightarrow 124 < 4n \Rightarrow n > 31 \).
So, the 32nd term is the first negative term.

Question. How many terms of AP 18, 16, 14, ... should be taken, so that their sum is zero?
Answer: \( a=18, d=-2, S_n = 0 \).
\( \frac{n}{2}[2(18) + (n-1)(-2)] = 0 \Rightarrow 36 - 2n + 2 = 0 \Rightarrow 2n = 38 \Rightarrow n = 19 \).

Question. Find the sum of first 8 multiples of 3.
Answer: AP: 3, 6, 9, ... (8 terms). \( a=3, d=3 \).
\( S_8 = \frac{8}{2}[2(3) + 7(3)] = 4[6 + 21] = 4 \times 27 = 108 \).

Question. Subha Rao started work in 1995 at an annual salary of ₹ 5000 and received an increment of ₹ 200 each year. In which year did his income reach ₹ 7000?
Answer: \( a = 5000, d = 200, a_n = 7000 \).
\( 5000 + (n-1)200 = 7000 \Rightarrow 200(n-1) = 2000 \Rightarrow n-1 = 10 \Rightarrow n = 11 \).
Year: \( 1995 + (11-1) = 2005 \).

Question. Ramkali saves ₹ 5 in the first week of a year and then increased her weekly savings by ₹ 1.75. If in the nth week, her weekly saving becomes ₹ 20.75. Find n.
Answer: \( a=5, d=1.75, a_n=20.75 \).
\( 5 + (n-1)1.75 = 20.75 \Rightarrow 1.75(n-1) = 15.75 \Rightarrow n-1 = \frac{15.75}{1.75} = 9 \Rightarrow n = 10 \).

Question. If \( \frac{1 + 3 + 5 + \dots \text{upto } n \text{ terms}}{2 + 5 + 8 + \dots \text{upto } 8 \text{ terms}} = 9 \), then find the value of n.
Answer: Numerator: sum of \( n \) odd numbers = \( n^2 \).
Denominator: sum of 8 terms of AP (2, 5, 8, ...), \( a=2, d=3 \).
\( S_8 = \frac{8}{2}[2(2) + 7(3)] = 4[4 + 21] = 100 \).
\( \frac{n^2}{100} = 9 \Rightarrow n^2 = 900 \Rightarrow n = 30 \).

Question. In an AP, if \( S_n = 3n^2 + 5n \) and \( a_k = 164 \), then find the value of k.
Answer: \( a_n = S_n - S_{n-1} = (3n^2+5n) - [3(n-1)^2+5(n-1)] = 6n + 2 \).
Given \( a_k = 164 \Rightarrow 6k + 2 = 164 \Rightarrow 6k = 162 \Rightarrow k = 27 \).

Question. Find the sum \( (-5) + (-8) + (-11) + \dots + (-230) \).
Answer: \( a = -5, d = -3, a_n = -230 \).
\( -5 + (n-1)(-3) = -230 \Rightarrow -3(n-1) = -225 \Rightarrow n-1 = 75 \Rightarrow n = 76 \).
\( S_{76} = \frac{76}{2}[a + l] = 38[-5 - 230] = 38 \times (-235) = -8930 \).

Question. Sum of the first n terms of an AP is \( 5n^2 - 3n \). Find the AP and also find its 16th term.
Answer: \( a_n = S_n - S_{n-1} = 10n - 8 \).
Terms: \( n=1 \rightarrow 2, n=2 \rightarrow 12, n=3 \rightarrow 22 \). AP: 2, 12, 22, ...
\( a_{16} = 10(16) - 8 = 152 \).

Question. The sum of the first n terms of an AP whose first term is 8 and the common difference is 20, is equal to the sum of first 2n terms of another AP whose first term is -30 and the common difference is 8. Find the value of n.
Answer: \( S_n = \frac{n}{2}[2(8) + (n-1)20] = n[8 + 10n - 10] = n(10n-2) \).
\( S'_{2n} = \frac{2n}{2}[2(-30) + (2n-1)8] = n[-60 + 16n - 8] = n(16n - 68) \).
\( n(10n-2) = n(16n-68) \Rightarrow 10n-2 = 16n-68 \Rightarrow 6n = 66 \Rightarrow n = 11 \).

Question. Find the sum of 10 terms of an AP. \( \sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \dots \)
Answer: AP: \( \sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, 4\sqrt{2}, \dots \). \( a = \sqrt{2}, d = \sqrt{2} \).
\( S_{10} = \frac{10}{2}[2\sqrt{2} + 9\sqrt{2}] = 5[11\sqrt{2}] = 55\sqrt{2} \).

Question. Find the sum of all multiples of 7 lying between 500 and 900.
Answer: First term > 500 divisible by 7 is 504. Last term < 900 is 896.
\( 504 + (n-1)7 = 896 \Rightarrow 7(n-1) = 392 \Rightarrow n-1 = 56 \Rightarrow n = 57 \).
\( S = \frac{57}{2}[504 + 896] = \frac{57}{2}[1400] = 57 \times 700 = 39900 \).

Question. Find the sum of all the two digit numbers which leave the remainder 2 when divided by 5.
Answer: Numbers: 12, 17, 22, ..., 97.
\( 12 + (n-1)5 = 97 \Rightarrow 5(n-1) = 85 \Rightarrow n-1 = 17 \Rightarrow n = 18 \).
\( S = \frac{18}{2}[12 + 97] = 9 \times 109 = 981 \).

Long Answer Type Questions

Question. If \( S_n \) denotes the sum of first n terms of an AP, then prove that \( S_{12} = 3 (S_8 - S_4) \).
Answer: \( RHS = 3 [\frac{8}{2}(2a+7d) - \frac{4}{2}(2a+3d)] = 3 [4(2a+7d) - 2(2a+3d)] = 3 [8a+28d - 4a-6d] = 3 [4a+22d] = 12a+66d \).
\( S_{12} = \frac{12}{2}[2a+11d] = 6(2a+11d) = 12a+66d \).
LHS = RHS. Proved.

Question. Find the sum of last ten terms of the AP 8, 10, 12, ..., 126.
Answer: Total terms: \( 8 + (n-1)2 = 126 \Rightarrow 2(n-1) = 118 \Rightarrow n-1 = 59 \Rightarrow n = 60 \).
Sum of last 10 = \( S_{60} - S_{50} \).
Alternatively, sum from end: \( a = 126, d = -2, n = 10 \).
\( S_{10} = \frac{10}{2}[2(126) + 9(-2)] = 5[252 - 18] = 5[234] = 1170 \).

Question. Find the sum of the first 100 natural numbers.
Answer: \( S = \frac{n(n+1)}{2} = \frac{100 \times 101}{2} = 50 \times 101 = 5050 \).

Question. Find the sum of first seven numbers which are multiples of 2 as well as of 9.
Answer: Multiples of 2 and 9 are multiples of 18.
AP: 18, 36, 54, ... (7 terms).
\( S_7 = \frac{7}{2}[2(18) + 6(18)] = \frac{7}{2}[36 + 108] = \frac{7}{2}[144] = 7 \times 72 = 504 \).

Question. For an AP, it is given that the first term \( a = 5 \), common difference \( d = 3 \), and the nth term \( a_n = 50 \). Find n and sum of first n terms \( S_n \) of the AP.
Answer: \( 5 + (n-1)3 = 50 \Rightarrow 3(n-1) = 45 \Rightarrow n-1 = 15 \Rightarrow n = 16 \).
\( S_{16} = \frac{16}{2}[5 + 50] = 8 \times 55 = 440 \).

Question. Find the sum of first 16 terms of an Arithmetic Progression whose 4th and 9th terms are -15 and -30, respectively.
Answer: \( a+3d = -15, a+8d = -30 \).
Subtracting: \( 5d = -15 \Rightarrow d = -3 \).
\( a + 3(-3) = -15 \Rightarrow a = -6 \).
\( S_{16} = \frac{16}{2}[2(-6) + 15(-3)] = 8[-12 - 45] = 8[-57] = -456 \).

Question. If the sum of first 14 terms of an Arithmetic Progression is 1050 and its fourth term is 40, find its 20th term.
Answer: \( S_{14} = 1050 \Rightarrow \frac{14}{2}[2a+13d] = 1050 \Rightarrow 2a+13d = 150 \).
\( a_4 = 40 \Rightarrow a+3d = 40 \Rightarrow 2a+6d = 80 \).
Subtracting: \( 7d = 70 \Rightarrow d = 10 \).
\( a = 40 - 3(10) = 10 \).
\( a_{20} = a+19d = 10 + 190 = 200 \).

Question. If the nth terms of the two AP’s 9, 7, 5, ... and 24, 21, 18, ... are the same, then find the value of n. Also, that term.
Answer: AP1: \( a=9, d=-2 \Rightarrow a_n = 9 + (n-1)(-2) = 11 - 2n \).
AP2: \( a=24, d=-3 \Rightarrow a_n = 24 + (n-1)(-3) = 27 - 3n \).
\( 11 - 2n = 27 - 3n \Rightarrow n = 16 \).
Term: \( a_{16} = 11 - 2(16) = -21 \).

Question. The 26th, 11th and the last terms of an AP are, 0, 3 and \( -\frac{1}{5} \), respectively. Find the common difference and the number of terms.
Answer: \( a+25d = 0, a+10d = 3 \).
Subtracting: \( 15d = -3 \Rightarrow d = -\frac{1}{5} \).
\( a = -25(-\frac{1}{5}) = 5 \).
Last term \( a_n = -\frac{1}{5} \Rightarrow 5 + (n-1)(-\frac{1}{5}) = -\frac{1}{5} \).
\( -\frac{1}{5}(n-1) = -5 - \frac{1}{5} = -\frac{26}{5} \Rightarrow n-1 = 26 \Rightarrow n = 27 \).

Question. The 4th term of an AP is zero. Prove that the 25th term of the AP is three times its 11th term.
Answer: \( a_4 = 0 \Rightarrow a+3d = 0 \Rightarrow a = -3d \).
\( a_{11} = a+10d = -3d+10d = 7d \).
\( a_{25} = a+24d = -3d+24d = 21d \).
Clearly \( a_{25} = 3 \times a_{11} \).

Question. If the mth term of an AP is \( \frac{1}{n} \) and nth term is \( \frac{1}{m} \), then show that its mnth term is 1.
Answer: \( a+(m-1)d = \frac{1}{n} \) and \( a+(n-1)d = \frac{1}{m} \).
Subtracting: \( (m-n)d = \frac{1}{n} - \frac{1}{m} = \frac{m-n}{mn} \Rightarrow d = \frac{1}{mn} \).
Substituting: \( a + \frac{m-1}{mn} = \frac{1}{n} \Rightarrow a = \frac{1}{n} - \frac{1}{n} + \frac{1}{mn} = \frac{1}{mn} \).
\( a_{mn} = a + (mn-1)d = \frac{1}{mn} + (mn-1)\frac{1}{mn} = \frac{1 + mn - 1}{mn} = 1 \).

Question. In an AP given that the first term \( a = 54 \), the common difference \( d = -3 \) and the n th term \( a_n = 0 \), find n and the sum of first n terms (\( S_n \)) of the AP.
Answer: \( 54 + (n-1)(-3) = 0 \Rightarrow -3(n-1) = -54 \Rightarrow n-1 = 18 \Rightarrow n = 19 \).
\( S_{19} = \frac{19}{2}[54 + 0] = \frac{19 \times 54}{2} = 19 \times 27 = 513 \).

Question. Solve \( 1 + 4 + 7 + 10 + \dots + x = 287 \).
Answer: AP: \( a=1, d=3 \). \( x = 1+(n-1)3 \).
\( S_n = \frac{n}{2}[1 + x] = 287 \Rightarrow \frac{n}{2}[2 + (n-1)3] = 287 \Rightarrow n(3n-1) = 574 \).
\( 3n^2 - n - 574 = 0 \). Using formula: \( n = \frac{1 \pm \sqrt{1 + 6888}}{6} = \frac{1+83}{6} = 14 \).
\( x = 1 + (13)3 = 40 \).

Question. Solve the equation: \( 1 + 5 + 9 + 13 + \dots + x = 1326 \)
Answer: \( a=1, d=4 \). \( \frac{n}{2}[2 + (n-1)4] = 1326 \Rightarrow n(1 + 2n - 2) = 1326 \Rightarrow 2n^2 - n - 1326 = 0 \).
\( n = \frac{1 \pm \sqrt{1 + 10608}}{4} = \frac{1+103}{4} = 26 \).
\( x = 1 + (25)4 = 101 \).

Question. Find the sum (i) \( 1 + (-2) + (-5) + (-8) + \dots + (-236) \) (ii) \( (4 - \frac{1}{n}) + (4 - \frac{2}{n}) + (4 - \frac{3}{n}) + \dots \) upto n terms.
Answer: (i) \( a=1, d=-3, a_n=-236 \Rightarrow 1+(n-1)(-3)=-236 \Rightarrow -3(n-1)=-237 \Rightarrow n=80 \).
\( S = \frac{80}{2}[1 - 236] = 40(-235) = -9400 \).
(ii) \( S = 4n - \frac{1}{n}(1+2+3+\dots+n) = 4n - \frac{1}{n}(\frac{n(n+1)}{2}) = 4n - \frac{n+1}{2} = \frac{8n-n-1}{2} = \frac{7n-1}{2} \).

Question. Find the sum of the two middle most terms of an AP \( -\frac{4}{3}, -1, -\frac{2}{3}, \dots, 4 \frac{1}{3} \).
Answer: \( a = -\frac{4}{3}, d = \frac{1}{3}, a_n = \frac{13}{3} \).
\( -\frac{4}{3} + (n-1)\frac{1}{3} = \frac{13}{3} \Rightarrow n-1 = 17 \Rightarrow n = 18 \).
Middle terms are \( \frac{18}{2} = 9\text{th} \) and \( 10\text{th} \).
\( a_9 = -\frac{4}{3} + 8(\frac{1}{3}) = \frac{4}{3} \).
\( a_{10} = -\frac{4}{3} + 9(\frac{1}{3}) = \frac{5}{3} \).
Sum = \( \frac{4}{3} + \frac{5}{3} = \frac{9}{3} = 3 \).

Question. Find the sum of first 17 terms of an AP whose 4th and 9th terms are -15 and -30, respectively.
Answer: As solved previously, \( a = -6, d = -3 \).
\( S_{17} = \frac{17}{2}[2(-6) + 16(-3)] = \frac{17}{2}[-12 - 48] = \frac{17}{2}(-60) = 17 \times (-30) = -510 \).

Question. The sum of first n terms of three AP’s are \( S_1, S_2, \) and \( S_3 \). The first term of each AP is unity and their common differences are 1, 2 and 3, respectively. Prove that \( S_1 + S_3 = 2 S_2 \).
Answer: \( S_1 = \frac{n}{2}[2 + (n-1)1] = \frac{n(n+1)}{2} \)
\( S_2 = \frac{n}{2}[2 + (n-1)2] = \frac{n(2n)}{2} = n^2 \)
\( S_3 = \frac{n}{2}[2 + (n-1)3] = \frac{n(3n-1)}{2} \)
\( S_1 + S_3 = \frac{n}{2}[n+1 + 3n-1] = \frac{n(4n)}{2} = 2n^2 = 2S_2 \). Proved.

Question. If the sum of first four terms of an AP is 40 and that of first 14 terms is 280. Find the sum of its first n terms.
Answer: \( \frac{4}{2}(2a+3d) = 40 \Rightarrow 2a+3d = 20 \).
\( \frac{14}{2}(2a+13d) = 280 \Rightarrow 2a+13d = 40 \).
Subtracting: \( 10d = 20 \Rightarrow d = 2 \).
\( 2a + 3(2) = 20 \Rightarrow 2a = 14 \Rightarrow a = 7 \).
\( S_n = \frac{n}{2}[2(7) + (n-1)2] = \frac{n}{2}[14 + 2n - 2] = n(n + 6) \).

Question. The ratio of the 11th term to the 18th term of an AP is 2 : 3. Find the ratio of the 5th term to the 21st term and also the ratio of the sum of the first five terms to the sum of the first 21 terms.
Answer: \( \frac{a+10d}{a+17d} = \frac{2}{3} \Rightarrow 3a+30d = 2a+34d \Rightarrow a = 4d \).
Ratio of \( a_5 : a_{21} = \frac{a+4d}{a+20d} = \frac{4d+4d}{4d+20d} = \frac{8d}{24d} = 1:3 \).
Ratio of \( S_5 : S_{21} = \frac{5/2[2a+4d]}{21/2[2a+20d]} = \frac{5[8d+4d]}{21[8d+20d]} = \frac{5 \times 12d}{21 \times 28d} = \frac{60}{588} = \frac{5}{49} \).

Question. The sum of four consecutive numbers in AP is 32 and the ratio of the product of the first and last terms to the product of two middle terms is 7 : 15. Find the numbers.
Answer: Let numbers be \( a-3d, a-d, a+d, a+3d \).
Sum = \( 4a = 32 \Rightarrow a = 8 \).
\( \frac{(a-3d)(a+3d)}{(a-d)(a+d)} = \frac{7}{15} \Rightarrow \frac{a^2-9d^2}{a^2-d^2} = \frac{7}{15} \Rightarrow 15a^2 - 135d^2 = 7a^2 - 7d^2 \).
\( 8a^2 = 128d^2 \Rightarrow 8(64) = 128d^2 \Rightarrow 512 = 128d^2 \Rightarrow d^2 = 4 \Rightarrow d = \pm 2 \).
Numbers: 2, 6, 10, 14.

Question. Show that the sum of an AP whose first term is a, the second term b and the last term c, is equal to \( \frac{(a + c)(b + c - 2a)}{2(b - a)} \).
Answer: \( a_1 = a, d = b-a, a_n = c \).
\( a + (n-1)(b-a) = c \Rightarrow n-1 = \frac{c-a}{b-a} \Rightarrow n = \frac{c-a+b-a}{b-a} = \frac{b+c-2a}{b-a} \).
\( S = \frac{n}{2}(a+c) = \frac{(b+c-2a)}{2(b-a)}(a+c) \). Proved.

Question. How many terms of the AP 20, \( 19 \frac{1}{3} \), \( 18 \frac{2}{3} \), ... must be taken, so that their sum is 300?
Answer: \( a=20, d=-\frac{2}{3}, S_n = 300 \).
\( 300 = \frac{n}{2}[40 + (n-1)(-\frac{2}{3})] \Rightarrow 600 = n[\frac{120 - 2n + 2}{3}] \Rightarrow 1800 = n(122 - 2n) \).
\( 2n^2 - 122n + 1800 = 0 \Rightarrow n^2 - 61n + 900 = 0 \).
Solving: \( (n-25)(n-36) = 0 \). \( n = 25 \) or \( n = 36 \).

Case Base Questions

Kanika was given her pocket money on Jan 1st, 2008. She puts ₹ 1 on day 1, ₹ 2 on day 2, ₹ 3 on day 3 and continued doing so till the end of the month. From this money into her piggy bank, she also spent ₹ 204 of her pocket money and found that at the end of the month she still had ₹ 100 with her.

Question. How much Kanika take till the end of the month from pocket money?
Answer: Jan 2008 has 31 days. Total savings = \( 1+2+3+\dots+31 \).
\( S = \frac{31(32)}{2} = 31 \times 16 = 496 \).
Total pocket money taken = Savings + Spent + Remaining = \( 496 + 204 + 100 = 800 \).

Question. How much was pocket money for the month?
Answer: Total pocket money = ₹ 800.

Question. What is the amount saved by Kanika, till January 13th, 2008?
Answer: Savings till 13th = \( 1+2+\dots+13 = \frac{13 \times 14}{2} = 91 \).
Amount saved = ₹ 91.